Question: Simplify the following expression and state the condition under which the simplification is valid. $k = \dfrac{x^2 - 9}{x + 3}$
First factor the polynomial in the numerator. The numerator is in the form ${a^2} - {b^2}$ , which is a difference of two squares so we can factor it as $({a} + {b})({a} - {b})$ $ a = x$ $ b = \sqrt{9} = 3$ So we can rewrite the expression as: $k = \dfrac{({x} + {3})({x} {-3})} {x + 3} $ We can divide the numerator and denominator by $(x + 3)$ on condition that $x \neq -3$ Therefore $k = x - 3; x \neq -3$